Question 762798
<pre>
Horner's method is simply synthetic division.

x³ + 5x² + 8x + 6 = 0

The feasible rational roots are ± the factors of 6, which are

±1, ±2, ±3, ±6

It has no sign changes (all terms positive), so there are
no positive roots.  So we need try only negative feasible
rational roots:

We try -1

-1|1  5  8  6 
  |<u>  -1 -4 -4</u> 
   1  4  4  2

No, that doesn't give 0 on the far right.

We try -2

-2|1  5  8  6 
  |<u>  -2 -6 -4</u> 
   1  3  2  2

No, that doesn't give 0 on the far right, either.

We try -3

-3|1  5  8  6 
  |<u>  -3 -6 -6</u> 
   1  2  2  0

Whoopie do! That give 0 on the far right!

So we have have found one zero -3. The above
synthetic division is a shortcut for long division
of dividing by (x + 3) and getting x² + 2x + 2 as a
quotient, so we have now factored:

   x³ + 5x² + 8x + 6 = 0

as

(x + 3)(x² + 2x + 2) = 0

Using the zero factor principle:

x + 3 = 0;       x² + 2x + 2 = 0
    x = -3       x = {{{(-b +- sqrt(b^2-4ac))/(2a)}}}
                 x = {{{(-(2) +- sqrt((2)^2-4(1)(2)))/(2(1))}}}             
                 x = {{{(-2 +- sqrt(4-8))/2}}}
                 x = {{{(-2 +- sqrt(-4))/2}}}
                 x = {{{(-2 +- 2i)/2}}}
                 x = {{{(2(-1 +- i))/2}}}
                 x = {{{(cross(2)(-1 +- i))/cross(2)}}}
                 x = -1 ± i

So the three roots are -3, -1 + i, -1 - i

Edwin</pre>